1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nadusha1986 [10]
3 years ago
15

Simplify 8(x-3)-4(2x-7)

Mathematics
1 answer:
Paraphin [41]3 years ago
4 0

Answer:

Your solution is 4.

Step-by-step explanation:

You might be interested in
Write a two point slope equations for the line passing through the points 6,5 and 3,1
Gwar [14]

The 2-point form of the equation of a line can be written as ...

... y = (y2-y1)/(x2-x1)·(x -x1) +y1

For your points, this is ...

... y = (1-5)/(3-6)·(x -6) +5

... y = (4/3)(x -6) +5

It can also be written as

... y -5 = (4/3)(x -6)

3 0
4 years ago
Read 2 more answers
What is the correct classification of the system of equations below? 14x + 2y = 10 y + 7x = -5 parallel coincident intersecting
marissa [1.9K]
In standard form the two equations are
  7x +y = 5
  7x +y = -5

The equations describe lines that are parallel.
5 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
Write an equation that is perpendicular to y= 1/2x+5 and passes through (4,8)
goblinko [34]

Answer:

y = -2x + 16.

Step-by-step explanation:

The slope of the perpendicular line = -1 / slope of the given line

= -1 / 1/2 = -2.

Using the point slope form of the equation of a straight line:

y - y1 = m (x - x1)

y - 8 = -2(x - 4)

y - 8 = -2x + 8

y = -2x + 16  is the required equation.

4 0
4 years ago
Do you like cake or ice cream more?
jenyasd209 [6]

Answer:

ice cream no no

Step-by-step explanation:

ice cream good i dont like the smell i am not bored

4 0
3 years ago
Read 2 more answers
Other questions:
  • What is 3/18 written in simplest form explain how you found you answer
    6·2 answers
  • The mean of 4 test scores is 90. What is the sum of the scores?
    8·2 answers
  • Consider that lines A and B are parallel. Which equation models the relationship between ∠1 and ∠5?
    12·2 answers
  • Tim's mom filled the gas tank in her SUV and spent $72.00 for 22.5 gallons of gasoline .what is the constant of proportionality
    12·1 answer
  • Help? Plz I need help
    10·1 answer
  • How many arches are in an equilateral triangle
    15·2 answers
  • 16.5−13.45+7.293 equals
    12·2 answers
  • Sue has 20 biscuits in a tin.
    5·1 answer
  • barbra has two cherry pies.She cuts each pie into quarters.How many slices of cherry pie does barbra have
    6·1 answer
  • Select all the points that are on the graph of the equation 4y - 6X equals 12
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!